ON ISOMETRIES OF EUCLIDEAN SPACES
F. S. BECKMAN AND D. A. QUARLES, JR.
1. Introduction. A transformation
of a Euclidean
which preserves unit distances is not necessarily an
is shown by the simple example of the transformation
all integral points one unit in the same direction and
line onto itself
isometry. This
which displaces
leaves all other
points fixed.
On the other hand, we show in this paper that a transformation of
Euclidean n-space En (2^»<«i)
which preserves a single nonzero
length must be a Euclidean (rigid) motion of En onto En. It is an immediate consequence that the result also holds for all finite-dimensional unitary spaces. For, if it were false for a unitary space of
dimension n, then it would be false for a Euclidean space of dimension 2m.
That this result fails for EK, the (real) Hubert space of infinite
sequences of real numbers {X} = {(xi, x2, • • • )} where ]C<" 1** exists, is shown by the following example.
There is in E°° a denumerable and everywhere dense set of points.
Let such a set be denoted by { 7<}. Define a transformation,
R, of
E™into { 7,} in such a way that the distance d(X, RX)<l/2.
Also,
define a transformation,
S, on {7<} so that SYi=Af where Ai
= (flu, a»2, • • •) and Oiy= 5iy21/s/2 (8,7 is the Kronecker delta). It may
be readily seen that T=SR
is a transformation
of Ex into itself which
preserves the unit distance. For, if d(Xi, X2) = l, then RXi¿¿RX2,
TXi^TXi,
and therefore d(TXu TX2) = l. However, not all distances
are preserved.
For, if Ai and A2 are any two points
in E°°,
then d(TXu TX2) is either 0 or 1.
2. Principal result.
Theorem. Let T be a transformation (possibly many-valued) of En
(2 g n < oo) into itself. Let d(p, q) be the distance between points p and
q of E", and let Tp, Tq be any images of p and q, respectively. If there
is a length a>0 such that d(Tp, Tq) =a whenever d(p, q) =a, then T is
a Euclidean transformation of En onto itself.
Remark.
By a change of scale we can assume in the proof that
o-l.
3. Proof of theorem for «=2.
Received by the editors December 3, 1952 and, in revised form, February
810
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21, 1953.
ON ISOMETRIESOF EUCLIDEANSPACES
811
Lemma 1. F is single-valued.
Proof. If F were not single-valued, there would be at least one
point, say pi, with at least two images, say Tp^ and Tpf\ Consider two points p2 and p3 which together with pi comprise the
vertices of an equilateral triangle with unit side. Since this kind of
triangular relationship must be preserved for each set of images of
pi, p2, and p3, it follows that Tp^ and Tpf\ together with an image
Tp2 of p2 and an image Tp3 of pz, comprise the vertices of a rhombus
of unit side with diTpP, Tpf))=31'2. Now consider two rhombi, each
congruent to the previously mentioned, whose vertices are pi, p2, pi,
pz and px, pi, pr, ps respectively, and where dipi, pi)=dipx, pj) = 3112
while dipi, pi) = l. Due to the relationship between p2, pz, and pi,
there must be an image, say Tpi, of pi which coincides with either
Tp[x) or Tp(2). Since images, say Tps and Tp6, oí pb and p6 must be
at a unit distance from each other and from both TpP and Tpf\ it
follows that the set {Tp6, Tpi} must coincide with the set { Tp2, Tpz}.
Thus, due to the relationship between ph, pit and p7, there must be
an image, say Tpi, of pi which coincides with either Tp[l) or Tpf\
implying that d(F/»4, Tpi) =0 or 31/2.The contradiction that dipi, p/)
= 1 is not preserved
d = 0 is preserved.
by F establishes
that
Lemma 2. If 31/2-l ^¿(¡pi, pt) áW'+í,
T is single-valued,
i.e.
then Tpx*Tp2.
Proof. It is clear that the vertices of an equilateral triangle with
unit side are transformed onto the vertices of a congruent triangle,
and that a unit circle is transformed into a congruent circle.
Consider the unit circles Cx and C2 with centers at px and p2. It is
possible to construct an equilateral
triangle of unit side with two
vertices on Cx and the third on C2. If Tpx = Tp2, then the three
vertices of an equilateral triangle with unit side would lie upon the
unit circle having Tpx = Tp2 as center, a contradiction.
Lemma 3. FAe distance 31/2 is preserved.
Proof. If a pair of vertices of a rhombus with unit sides are separated by a distance 31/2 they must, from Lemma 1, be transformed
into a single point or onto two points which are separated by a distance 31/2. Lemma 2 implies that the distance 31/2 is'preserved.
That the distance 2 is preserved is seen by considering the regular
hexagon with unit side. The consecutive vertices of such a figure
must be transformed consecutively onto the vertices of a congruent
figure. This becomes apparent when one considers the component
overlapping rhombi. Therefore, two diametrically opposite vertices
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812
F. S. BECKMANAND D. A. QUARLES
[October
(which are separated by a distance 2) are transformed onto points
separated by the same distance. The extension of such overlapping
rhombi to establish the preservation of any integral distance is obvious.
We shall now show that all distances of the form A/2B, where
A and B are integers, are preserved. By an obvious extension of the
proof of Lemma 2, it follows that if A(31'2-l)
^d(pu p2) uA(3"2+l)
then Tpi7¿Tp2. Consider the isosceles triangle with two sides of
length 2^4 and the third of length A. The vertices of this triangle are
transformed onto the vertices of a congruent triangle. Thus, the system of two intersecting circles, each of which has one of the sides of
length 2.4 as a diameter, is transformed
into a congruent system
since a circle of radius A is transformed into a congruent circle. The
point, say pi, of intersection of the two circles which is coincident
with a vertex of the isosceles triangle is transformed into the corresponding point of the congruent image system. The other point,
say pi, of intersection (which is the midpoint of the side of length A)
must be transformed into a point of intersection of the circles of the
congruent image system. Since A(3xl2 —i)<d(pi,
p2) = (A/2)(l5)112
•C4(31/l+l),
¡t follows that Tpu¿Tp2. Thus, the midpoint of the
side of length A is transformed into the midpoint of the line joining
the images of its end points, and the distance A/2 is preserved. By
extension of this argument, the distances .4/22 and hence A/2B are
preserved.
Since the points .4/2s are dense on the non-negative
portion of
the real line, there can be constructed, corresponding to arbitrary
points p and q of the plane, a sequence of points {qi}, with each
d(p, qi) of the form A/2B, where lim,-,„d(p, qi)=d(p,
q), and where
lim,_wg, = g. For, if qi is sufficiently close to q, both q( and q will lie
on a circle of arbitrarily small radius 1/2B. It follows that Tqi and
Tq lie on a circle of the same radius, and therefore limi^KTqi—Tq.
Also, limi^xd(Tp, Tqi) =d(p, q), since, for all i, d(p, qi) is preserved.
Therefore, d(Tp, Tq)=d(p, q), and hence T preserves all distances.
4. Proof of theorem for 3^»< °o. Generalization of Lemmas 1,2,
and 3. That T is single-valued may be shown by considering an
(n —l)-dimensional unit-edged simplex in «-space. To the n vertices
of this figure there can be adjoined in only two ways an additional
vertex in such a way as to form an «-dimensional unit-edged simplex.
By analogous reasoning to that used in proving Lemma 1, the assumption that some point has more than one image leads to the contradiction that the distance between the images of two unit-distant
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19531
ON ISOMETRIESOF EUCLIDEANSPACES
813
points is either 0 or [2(»+l)/»]1/2
(i.e., 0 or twice the altitude of an
«-dimensional unit-edged simplex).
Clearly, the vertices of a A-dimensional (A^w.) unit-edged simplex
in »-space are transformed by T onto the vertices of a congruent
simplex, and the points on an »-dimensional hypersphere of unit
radius are transformed into a congruent hypersurface. As a generalization of Lemma 2, if [2(« + l)/»]1/2-l
gdipt, pi) = [2(m+ 1)/»]1/2
+ 1, then TpxT^Tpi. For, otherwise, there would exist an »-dimensional unit-edged simplex in «-space whose vertices have images
which both comprise the vertices of a congruent simplex and yet lie
upon an «-dimensional hypersphere of unit radius, a contradiction.
(The edge of an »-dimensional equi-edged simplex inscribed in an
»-dimensional
hypersphere of unit radius is easily shown to be
[2(«+l)/«]1'2>l.)
The distances [2in + l)/n]112, .[2(w+l)/w], • • • , [2(«+l)/»]'/2,
• • • are preserved. As in the proof of Lemma 3, it follows from the
above generalizations of Lemma 1 and Lemma 2 that the distance
[2(« + l)/«]1/2 is preserved. Replacing the unit distance by the distance [2(»+l)/»]1/2,
it then follows that [2(m+ 1)/m] is preserved,
and thence [2(« + l)/«]3/2 is preserved, etc.
Lemma 4. A unit circle and its center are transformed into a unit
circle and its center, respectively.
Proof. For simplicity of exposition, the case « = 3 will be treated
separately.
By the generalization of Lemma 3, any spherical surface of radius
(2/3)61/2 is transformed into a congruent surface. Two such spherical
surfaces whose centers are separated by a distance (2/3)(15)1/2 inter-
sect in a unit circle. Since (2/3)61'2-l<
(2/3) (15) 1'2<(2/3)61/2+l,
it
follows from the generalization of Lemma 2 that this unit circle is
transformed into the intersection of two distinct spherical surfaces.
Thus, any unit circle in 3-space is transformed into a circle.
Now consider the vertices of a regular hexagon px, p2, • • • , pt inscribed in a unit circle Cx. Since Tpi^Tpj
(tV/) by the generalization
of Lemma 2, and since dipx, p2)=dip2, pz)= • • • =¿(p6, pi) = l, it
follows that Tpx, Tp2, • • • , Tp& on FCi are the vertices of a regular
hexagon with unit side. Therefore, Cx is transformed into a unit circle,
and the image circle's center, which is the only point in space at unit
distance from Tpx, Tp2, • • • , Tp6, must be the image of the center
of Cx.
The case 4^«<
oo will now be treated.
It will first be shown that, if s is a preserved
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distance,
then the
814
F. S. BECKMAN AND D. A. QUARLES
[October
surface of any 3-dimensional sphere of radius s[(n —l)/2(n —2)]1/2 in
«-space is transformed into a congruent surface. Let coordinate axes
in «-space be designated by Xi, A2, • • • , Xn. Consider an (« —3)dimensional equi-edged simplex (edge = s) which is situated with its
centroid at the origin, and in the intersection of hyperplanes Xi = x2
= x3 = 0. Let the vertices of this simplex be designated by (0, 0, 0,
ci\
c&\ • • " i cn'). i = l. 2, • • • , » —2. Let « —2 «-dimensional
hyperspheres of radius s be located so that each of the vertices of the
simplex is a center of one hypersphere. The equations of these hyperspherical
surfaces
are then:
x\+x\-\-x\-\-(xi
—c^)2-^ ■ • ■ +(xn
—Cn{i))2= s2,j = l, 2, • • ■ ,« —2. By subtraction, we obtain from these
equations
the
equations:
(¿P —c^')x4 + • • • + (41' —c„ )xn = 0,
j = 2, 3, • • • , « —2. Since the determinant of the array of coefficients
of this last set of equations is nonzero, it follows that xt = x6 = • • •
=x„ = 0. Since Zt=*(ct1))2 = (n —3)s2/2(n —2), the common intersection of the « —2 hyperspherical surfaces is expressed by the equations:
x?+x^+x^ = (m—l)s2/2(w —2) and X4= xs= • • • =x„ = 0. Thus, the
surface of any 3-dimensional sphere of radius s[(n —l)/2(n —2)]1/2
can be considered to be the intersection of such a system of « —2
hyperspherical
surfaces of radius s, when the origin of the coordinate
system is taken at the center of the 3-dimensional sphere. The image
of this system of hyperspherical surfaces lies in a congruent system,
and therefore the surface of the 3-dimensional sphere is transformed
into the surface of a congruent 3-dimensional sphere.
It will now be shown that any unit circle in «-space is transformed
into a circle. Any unit circle can be considered to be the intersection
of the surfaces of two 3-dimensional spheres of radii [2(« + l)/«]I/2
• [(«-l)/2(«-2)]1'2
and [2(n+l)/n][(n-l)/2(n-2)]1'2.
By the
preceding paragraph, these spherical surfaces are transformed
into
surfaces which are congruent to them. These, consequently distinct,
spherical image surfaces contain the image of the unit circle in their
circle of intersection.
The completion of the proof of Lemma 4 for 4 ^ » < °o follows by
the second paragraph of the argument for the case « = 3.
Lemma 5. A plane is transformed into a plane.
Proof. Let G be a unit circle in «-space with center ou and let p
be an arbitrary point in the plane of G such that d(p, Oi) ?S2. Construct a co-planar unit circle, G, passing through p with its center on
the circumference of &. By Lemma 4, G and G are transformed into
intersecting unit circles. Furthermore,
the images of the two points
of intersection of the circles G and G and the image of oi can be seen
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19531
ON ISOMETRIES OF EUCLIDEAN SPACES
815
to determine the plane of both FCi and TC2. Therefore, Tp lies in this
plane and all points in the plane of Cx within a distance 2 from Oxare
transformed
into co-planar
points.
Obviously,
this region can be ex-
tended in the plane of Cx indefinitely, since Cx could now, in the
above argument, be chosen to be a unit circle which is internally
tangent to the boundary of this region.
Since planes are transformed into planes, it follows, from the proof
of the theorem for the case » = 2, that all distances are preserved.
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